This result is a consequence of some musings in the last week. A few basic results will need to be proven first. The main result is:

Or more generally we have the following corollary for polynomials with integer terms:

I scribbled this Proposition (below) last week and wondered if it’s true in general. The proof turned out rather simple. I like this fact because of its general nature.

The proof proceeds by showing first that given some integer of some parity, its powers remain of the same parity. For even, it follows immediately, for odd a simple argument using the

**Example**. The above result guarantees**for any***x***∊ ℤ a solution such that**in the following equations:*y*∊ ℤ*y*=

**½**(

*a*

*x*

^{3}+

*b*

*x*

^{12}+

*c*

*x*

^{81}+

*d*

*x*

^{739}+

*e*

*x*

^{993}- (

*a + b + c + d + e*)

*x*)

*y*=

**½**(

*a*

_{1}

*x*

^{1}+

*a*

_{2}

*x*

^{2}+

*a*

_{3}

*x*

^{3}+ ... +

*a*

_{n}*x*

^{179}- (Σ

*a*

*)*

_{i}*x*)

Or more generally we have the following corollary for polynomials with integer terms:

I scribbled this Proposition (below) last week and wondered if it’s true in general. The proof turned out rather simple. I like this fact because of its general nature.

**Proposition 1***For**n*∊ℤ^{+},*k*,*x*∊ℤ, and*m*∊ℕ.*x*+

^{n}*x*

^{n+m}= 2

*k*

The proof proceeds by showing first that given some integer of some parity, its powers remain of the same parity. For even, it follows immediately, for odd a simple argument using the

*Binomial Theorem*shows that the remaining constant in an arbitrary expansion of 2*k*+ 1 is 1, and the remining terms factorise as multiples of 2 i.e. (2*k*+ 1)*= 2*^{n}*m*+1 m∊ℤ.^{The result from the top of the page is a direct corollary of Proposition 1.}
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